Derivative of an Odd Function

Date: 02/27/2002 at 00:22:13
From: Sheri
Subject: Derivative of an odd function is even
How do you show that the derivative of an odd function is even?
That is, if f(-x) = -f(x), then f'(-x) = f'(x).
I have been working on this problem for two days and still can't
figure out the answer.

Date: 03/01/2002 at 01:27:42
From: Doctor Douglas
Subject: Re: Derivative of an odd function is even
Hi, Sheri,
Thanks for submitting your question to the Math Forum.
We start from the definition of the derivative:
f'(x) = lim{h->0} [f(x+h) - f(x)]/h
Note that this must be defined for all h sufficiently close to zero.
In particular, the limit must exist whether h approaches zero from the
right or from the left (and these limits must be equal to one another,
of course).
Now let us consider f'(-x) for a function that is odd.
f'(-x) = lim{h->0} [f(-x+h) - f(-x)]/h now let g = -h
= lim{g->0} [f(-x-g) - f(-x)]/(-g) now use "f is odd" twice
= lim{g->0} [-f(x+g) - (-f(x))]/-g the rest is just
manipulation...
= lim{g->0} [f(x+g) - f(x)]/g
= f'(x) by def'n of derivative
Thus, if an odd function possesses a derivative, it is even.
- Doctor Douglas, The Math Forum
http://mathforum.org/dr.math/